of 1 42
The Elements, Proton-Seconds, And Quanta (Short)
Ian Beardsley!
(University of Oregon, Department of Physics, 2022)!
!
Genesis Project California 2022!
⋅
⋅
of 2 42
Abstract: It would seem to not only are artificial intelligence elements mathematical constructs,
but so are biological life elements, and they are described in terms of one another, this is the
subject of part 1. It becomes necessary to say if these elements are subject to mathematical
constructs then matter, or inertia is a mathematical construct. This results in a constant that not
only predicts the hydrocarbons, the skeletons of life, but determines a definition for the radius
of the solar system and predicts the radius of the proton. This is part 2. In part 3 we show this
takes the form of a gaussian distribution meaning it can all be solved with wave mechanics.
This leads to predicting the relative abundances of hydrogen and helium from which the other
elements are made by stars in nucleosynthesis."
of 3 42
Part 1 The Mathematical Construct…………….4!
Part 2 The Constant………………………………10!
Part 3 The Wave Packet………………………….33!
Part 4 Where It Began……………………………42!
Part 5 Archeology of other Star Systems………50"
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Part 1: The Mathematical Construct!
of 5 42
Building A Matrix !
We pull these AI elements out of the periodic table of the elements to make an AI periodic
table:!
We now notice we can make a 3 by 3 matrix of it, which lends itself to to the curl of a vector
field by including biological elements carbon C (above Si):!
=!
=!
!
Which resulted in Stokes theorem (Beardsley, Essays In Cosmic Archaeology Volume 3):!
Equation 1. !
Where!
!
!
!
We were then able to write this with product notation!
i
j
k
∂
∂x
∂
∂y
∂
∂z
(−C ⋅ P)y (Si ⋅ Ga)z (Ge ⋅ As)y
(Ge ⋅ As − Si ⋅ G a)
i + (C ⋅ P)
k
[
(72.64)(74.92) − (28.09)(69.72)
]
i +
[
(12.01)(30.97)
]
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a = ex p
(
1
Ge − Si
∫
Ge
Si
ln(x)d x
)
∇ ×
u = (Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
d
a =
(
zdydz
i + yd ydz
k
)
u = (−C ⋅ P)y
i + (Si ⋅ Ge)z
j + (Ga ⋅ As)y
k
of 6 42
Equation 2. !
While we have the AI BioMatrix!
Which we used to formulate a similar equation (Beardsley, Essays In Cosmic Archaeology
Volume 2)!
We can form another 3X3 matrix we will call the electronics matrix (Beardsley, Cosmic
Archaeology, Volume Three):!
!
We can remove the 5th root sign in the above equation by noticing!
Equation 3. !
=(28.085)(72.64)(12.085)(107.8682)(196.9657)=!
!
Where we have substituted carbon (C=12.01) the core biological element for copper (Cu).!
But since we have:!
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a =
n
n
∏
i=1
x
i
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ Cu ⋅ Ag ⋅ Au
523,818,646.5
g
5
mol
5
of 7 42
Equation 4. !
We take the ratio and have!
!
Almost exactly 3 which is the ratio of the perimeter of regular hexagon to its diameter used to
estimate pi in ancient times by inscribing it in a circle:!
!
Perimeter=6!
Diameter=2!
6/2=3!
!
Thus we have the following equation…!
Equation 5. !
Showing The Calculation using the most accurate data possible…!
Ge=72.64!
As=74.9216!
Si=28.085!
Ga=69.723!
C=12.011!
P=30.97376200!
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u ) ⋅ d
a = 170,535,359.662(g/m ol )
5
523,818,646.5
170,535,359.662
= 3.0716
π = 3.141...
π
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u ) ⋅ d
a =
5
∏
i=1
x
i
of 8 42
=!
=!
!
!
!
!
!
=154,082,837.980+16,452,521.6822=!
!
=!
(28.085)(72.64)(12.085)(107.8682)(196.9657)=!
!
Where we have substituted carbon C=12.01 for copper Cu. We use Cu, Ag, Au because they
are the middle column of our electronics matrix, they are the finest conductors used for
electrical wire. We use C, Si, Ge because they are the middle column of our AI Biomatrix. Si
and Ge are the primary semiconductor elements used in transistor technology (Artificial
Intelligence) and C is the core element of biological life. We have!
!
(Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.9216) − (28.085)(69.723)
]
i +
[
(12.011)(30.97376200)
]
k
3,484.134569
(
g
mol
)
2
i + 372.025855
(
g
mol
)
2
k
∫
Ge
Si
∫
Ge
Si
(
3,484.134569
(
g
mol
)
2
i + 372.025855
(
g
mol
)
2
k
)
⋅
(
zdydz
i + yd ydz
k
)
∫
Ge
Si
∫
Ge
Si
(
3,484.134569
(
g
mol
)
2
⋅ zdzdy + 372.025855
(
g
mol
)
2
⋅ yd zd y
)
∫
Ge
Si
3,484.134569
(
(72.64 − 28.085)
2
2
)
dy +
∫
Ge
Si
372.025855y ⋅ (72.64 − 28.085)d y
3458261.42924
(
g
mol
)
4
(72.64 − 28.085) + 16575.6119695
(
g
mol
)
3
(
(72.64 − 28.085)
2
2
)
170,535,359.662
(
g
mol
)
5
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ Ag ⋅ Au
523,818,646.5
g
5
mol
5
523,818,646.5
170,535,359.662
= 3.0716
of 9 42
!
Perimeter/Diameter of regular hexagon = 3.00!
!
The same value as our 3.0716 if taken at two places after the decimal."
π = 3.141...
3.141 + 3.00
2
= 3.0705
of 10 42
Part 2: The Constant!
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Introduction
In order to present the elements as mathematical structures we need to explain the matter from
which they are made as mathematical constructs. We need a theory for Inertia. I had found
(Beardsley Essays In Cosmic Archaeology. 2021) where I suggested the idea of proton
seconds, that is six proton-seconds, which is carbon the core element of biological life if we
can figure out a reason to divide out the seconds. I found!
Equation 1. !
Where h is Planck’s constant 6.62607E—34 Js, is the radius of a proton 0.833E-15m, G is
the universal constant of gravitation 6.67408E-11 (Nm2)/(kg2), and c is the speed of light
299,792,459 m/s. And is t=1 second. is the Sommerfeld constant (or fine structure
constant) is 1/137. The mass of a proton is .!
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared: !
!
Matter is that which has inertia. This means it resists change in position with a force applied to
it. The more of it, the more it resists a force. We understand this from experience, but what is
matter that it has inertia?!
I would like to answer this by considering matter in one of its simplest manifestations, the
proton, a small sphere with a mass of 1.6726E-27 kg. This is a measure of its inertia.!
I would like to suggest that matter, often a collection of these protons, is the three dimensional
cross-section of a four dimensional hypersphere.!
The way to visualize this is to take space as a two-dimensional plane and the proton as a two
dimensional cross-section of a sphere, which would be a circle.!
In this analogy we are suggesting a proton is a three dimensional bubble embedded in a two
dimensional plane. As such there has to be a normal vector holding the higher dimensional
sphere in a lower dimensional space. Thus if we apply a force to to the cross-section of the
sphere in the plane there should be a force countering it proportional to the normal holding it in
a lower dimensional universe. This counter force would be experienced as inertia. It may even
induce in it an electric field, and we can see how it may do the same equal but opposite for the
electron. Refer to the illustration on the following page…"
1
t
1
α
2
m
p
h4π r
2
p
Gc
= 6protons
r
p
t
1
α
m
p
= 1.67262E − 27kg
α
2
=
U
e
m
e
c
2
of 12 42
"
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!
!
!
I made a program that looks for close to whole number solutions, I set it at decimal part equal
to 0.25. You can choose how may values for t you want to try, and by what to increment them.
Here are the results for incrementing by 0.25 seconds then 0.05 seconds. Constant to all of this
is hydrogen and carbon. The smaller integer value of seconds gives carbon (6 protons at 1
second) and the largest integer value of seconds gives hydrogen (1 proton at six seconds) and
outside of that for the other integer values of protons you get are at t>0 and t<1. Equation 1
really has some interesting properties. Here are two runs of the program( decpart is just me
verifying that my boolean test was working right to sort out whole number solutions):!
By what value would you like to increment?: 0.25!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
By what value would you like to increment?: 0.05!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
40.1998 protons 0.150000 seconds 0.199837 decpart !
30.1499 protons 0.200000 seconds 0.149879 decpart !
24.1199 protons 0.250000 seconds 0.119904 decpart !
20.0999 protons 0.300000 seconds 0.099918 decpart !
17.2285 protons 0.350000 seconds 0.228500 decpart !
15.0749 protons 0.400000 seconds 0.074938 decpart !
12.0599 protons 0.500000 seconds 0.059950 decpart !
10.0500 protons 0.600000 seconds 0.049958 decpart !
8.0400 protons 0.750000 seconds 0.039966 decpart !
1
α
2
m
p
h4π r
2
p
Gc
= (6protons)(1secon d )
1
α
2
m
p
h4π r
2
p
Gc
= (1proton)(6secon ds)
α
2
=
U
e
m
e
c
2
of 14 42
7.0941 protons 0.850000 seconds 0.094088 decpart !
6.0300 protons 1.000000 seconds 0.029975 decpart !
5.2435 protons 1.150000 seconds 0.243457 decpart !
5.0250 protons 1.200000 seconds 0.024980 decpart !
4.1586 protons 1.450000 seconds 0.158605 decpart !
4.0200 protons 1.500000 seconds 0.019985 decpart !
3.1737 protons 1.899999 seconds 0.173673 decpart !
3.0923 protons 1.949999 seconds 0.092296 decpart !
3.0150 protons 1.999999 seconds 0.014989 decpart !
2.2333 protons 2.699999 seconds 0.233325 decpart !
2.1927 protons 2.749999 seconds 0.192719 decpart !
2.1536 protons 2.799999 seconds 0.153564 decpart !
2.1158 protons 2.849998 seconds 0.115782 decpart !
2.0793 protons 2.899998 seconds 0.079303 decpart !
2.0441 protons 2.949998 seconds 0.044061 decpart !
2.0100 protons 2.999998 seconds 0.009993 decpart !
1.2433 protons 4.850000 seconds 0.243294 decpart !
1.2306 protons 4.900001 seconds 0.230607 decpart !
1.2182 protons 4.950001 seconds 0.218177 decpart !
Here is the code for the program:!
#include <stdio.h>!
#include <math.h>!
int main(int argc, const char * argv[]) {!
!
int n;!
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;!
!
do!
{!
printf("By what value would you like to increment?: ");!
scanf("%f", &increment);!
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");!
scanf("%i", &n);!
}!
while (n>=101);!
{!
for (int i=0; i<n;i++)!
{!
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));!
!
int intpart=(int)protons[i];!
float decpart=protons[i]-intpart;!
t=t+increment;!
if (decpart<0.25)!
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);!
}}}}!
of 15 42
While I understand that one second is a human invention and can’t be taken as significant, the
equations have meaning in that there are two equations each utilizing a second so they are
connected. But what really makes me wonder is how one can predict carbon, and the other
hydrogen so accurately with the unit of a second. It was a conspiracy on the part of those who
formulated the duration of a second a long time ago to be what it is? The Ancient Egyptians,
The Babylonians, The Julian calendar—who, what, when?!
6 protons gives a little more than a second. This makes a shorter day. We have!
Equation 2. !
H=1.00784 g/mol, carbon = 6 protons!
h=6.62607E-34, r_p=0.833E-15, G=6.67408E-11, c=299,792,459!
The Second
But actually the second might have physical meaning beyond what is here. Not only is the
second related to the earth orbital period but, to other things; By dividing the day into 24 hours,
the hour into 60 minutes, and the minute into 60 seconds, the second is 1/86400 of day. By
doing this we have a twelve-hour daytime at spring and fall equinox on the equator, 12 being
the most divisible number for its size (smallest abundant number). That is to say that twelve is
evenly divisible by 1,2,3,4,6 which precede it and 1+2+3+4+6=16 is greater than twelve. As
such there is about one moon per 30 days and about 12 moons per year (per each orbit) giving
us a twelve-month calendar. This is all further convenient in that the moon and earth are in very
close to circular orbits and the circle is evenly divisible by 30, 45, 60, and 120 if we divide the
circle into 360 degrees which are special angles very useful to the workings of physics and
geometry. Further, the 360 degrees of a circle are about the 365 days of a year (period of one
earth orbit) so as such the earth moves through about a degree a day in its journey around the
sun. Thus, through these observations down through the ages since ancient times we have
constructed the duration of a second wisely enough to make a lot work together. Now we see 6
protons, which is carbon the core element of biological life on the planet where all of this came
together is deeply connected with the second that defines it all. With this idea of proton-
seconds describing hydrogen and carbon the basis of life the hydrocarbons in a cycle of 6 with
respect to one another, the motion of the earth around the sun and moon around the earth, and
the basis of geometry the 360 degree circle, equation 1 connects them with the universal
constant of gravitation, the speed of light, the fine structure constant and Planck’s constant
that characterize the physics of the atom. I really wonder if other star systems are connected
so well to their planets with their star as is the Earth with the Sun. We really can’t resolve
planets around other stars because the stars are too far and so bright compared to their
orbiting planets and the planets have to be large enough to be inferred by the motion they
induce in the star they orbit. It is hard to do so with earth sized planets that might harbor life.!
The Geometrical Explanation of Seconds
Here we will talk about the equation 1:!
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon ds
1
t
1
α
2
m
p
h4π r
2
p
Gc
= 6protons
of 16 42
In so far as the second unifies carbon (6 protons) with hydrogen (1 proton) through the unit of a
second as the hydrocarbons the backbones of life. We have suggested the second is important
as well in terms of the phases of the moon and the earth and that these determined the
calendar system we use. We further suggested there is a connection of this to the structure
found in geometry, and this is what we want to explore further, here. We ended with all of this
can be compactly written as:!
!
Where n=3, 4, 5, ,6 ,…!
We could evaluate this for n equal to other integers, or even the numbers, but these produce
the special triangles, and geometries we are most interested. Thus we will begin by pointing
out that!
3 X 4 X 5 X 6 = 360!
The amount of degrees into which we divide a circle and that, as such it approximates the
number of days in a year (1 revolution of the earth around the sun) and thus we see that!
Equation 3. !
Represents days as well (The earth moves through about 1 degree a day in its orbit around the
sun) by solving it for n:!
Equation 4. !
Where,!
!
Which correspond respectively to:!
!
Which are the unit triangle, unit the square, the unit regular pentagon, and the unit regular
hexagon.!
The Derivative
We take our function!
And write it:!
!
Then take the derivative:!
2 cos(π /n) = 1, 2, Φ, 3, . . .
2 cos(π /n) = 1, 2, Φ, 3, . . .
days = cos
−1
(y/2)
y = 1, 2,
5 + 1
2
, 3, . . .
n = 3,4,5,6,...
days = cos
−1
(y/2)
y = cos
−1
(x /2)
of 17 42
Equation 5. !
This is a right triangle with hypotenuse 2, and height x and with base :!
To get , then . To get , then , and to get then .
To get then . Thus our function is equation four written:!
!
Where:!
!
n=3,4,5,6,,…!
=!
- is the rate of change of a triangle (x=1). -0.85 is the rate of change of a regular
pentagon ( ). - is the rate of change of a square ( ). -1 is the rate of
change of a regular hexagon ( ).!
We see that!
!
!
Is x as a function of n, and that n=3 is a unit triangle, n=4 a unit square, n=5 is a unit pentagon,
and n=6 is a unit hexagon. Thus if n=3 we have the unit triangle is Earth, the unit square is
dy
dx
= −
1
4 − x
2
4 − x
2
θ = 60
∘
x = 3
θ = 45
∘
x = 2
θ = 30
∘
x = 1
θ = 36
∘
4 − x
2
2
= Φ
y = 2cos(θ(n))
θ(n) =
π
n
=
180
∘
n
dy
dx
= −
1
4 − x
2
1/ 3
f : x ↦ Φ
2
2
x = 2
x = 3
y = cos
−1
(x /2)
dy
dx
= −
1
4 − x
2
of 18 42
Mars, the unit pentagon is n=5 which not only is this shape not a member of the regular
tessellators, n=5 is the asteroid belt, which is a location in the solar system where a planet
cannot form. We then proceed to Jupiter, which is n=6, the most massive planet in the solar
system which carries the majority of its mass. dy/dx is the the change in days with respect to
planetary number.!
The Formulation
We can actually formulate this differently than we have. We had!
!
!
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2
are lower and upper limits in an integral, then we have:!
Equation 6. !
And we can do this:!
!
If we take our equation 4 and integrate over the same range:!
!
If we add these two the result we get is:!
0.7806+0.21039=0.99~1.00!
1
t
1
1
α
2
m
p
h4π r
2
p
Gc
= 6protons
1
t
6
1
α
2
m
p
h4π r
2
p
Gc
= 1proton
1
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.7806
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21039
of 19 42
The Solar System And Sand
While we have!
!
Is a number of protons!
!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure
number. We have that!
!
!
Interestingly 78% is the percent of N2 in the atmosphere and 21% is the percent of O2 in the
atmosphere (by volume). These are the primary constituents that make it up. The rest is
primarily argon and CO2. This gives the molar mass of air as a mixture is:!
!
Now interestingly, I have found this connected to the solar system by considering a mixture of
silicon, the primary constituent of the Earth crust, and germanium just below it in the periodic
table, in the same proportions of 78% and 21%. Silicon is also the primary second generation
semiconductor material (what we use today) and germanium is the primary first generation
semiconductor material (what we used first). The silicon is directly below our carbon of one
proton-second, silicon directly below that, and germanium directly below that, in the periodic
table. So we are moving directly down the periodic table in group 14. The density of silicon is
2.33 g/cm3 and that of germanium is 5.323 g/cm3. Let us weight these densities with our 0.21
and 0.78:!
!
Now consider this the starting point for density of a thin disc decreasing linearly from the Sun
to Pluto (49.5AU=7.4E14cm). Thus,…!
!
1
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
1
α
2
m
p
h4π r
2
p
Gc
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
0.78N 2 + 0.21O 2 = 29.0g/m ol
0.21Si + 0.78Ge = 4.64124g /cm
3
ρ(r) = ρ
0
(
1 −
r
R
)
of 20 42
Thus,…!
!
Or,…!
!
Thus,…!
!
If we add up the masses of the planets in our solar system they are 2.668E30 grams.!
Since!
!
Meaning mixing germanium and silicon in the same proportion that occurs with N2 and O2 in
the atmosphere and with!
!
!
Where!
!
In the first integral. See the following pages to see the computation of the mass of the planets
in the solar system…"
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
rdrdθ
M =
πρ
0
R
2
3
M =
π(4.64124)(7.4E14)
2
3
= 2.661E 30
2.661
2.668
(100) = 99.736
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
6 =
1
α
2
m
p
h4π r
2
p
Gc
of 21 42
"
of 22 42
As we can see Jupiter carries the majority of the mass, Saturn a pretty large piece, and
somewhat large Uranus and Neptune. We don’t even list Pluto’s mass. When we consider!
!
!
We are considering and . These come from!
!
!
From 30 degrees to 45 degrees is 15 degrees. The Earth rotates through 360/24 is 15 degrees
per hour. The hour is divided into 60 minutes and minutes into 60 seconds…(Next page)"
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
2
3
2 cos(45
∘
) = 2
2 cos(30
∘
) = 3
of 23 42
Radius of the Solar System
We have said!
!
For a thin disc. Thus we have a definition for the radius of the solar system, :!
Equation 7. !
Where!
Equation 8. !
Equation 9. !
Equation 10. !
Equation 11. !
is the mass of all the planets. We also have that the molar mass of air to the molar mass of
water is approximately the golden ratio. The interesting thing is we determine a definition for
the radius of the solar system and predict the hydrocarbons (backbones of life) all in one fell
swoop.!
Proton Planck-Seconds
By what value would you like to increment?: 0.01
How many values would you like to calculate for t in equation 1 (no
more than 100?): 100
86.1425 protons 0.070000 seconds 0.142517 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
43.0713 protons 0.140000 seconds 0.071259 decpart
40.1998 protons 0.150000 seconds 0.199841 decpart
30.1499 protons 0.200000 seconds 0.149876 decpart
26.2173 protons 0.230000 seconds 0.217283 decpart
25.1249 protons 0.240000 seconds 0.124893 decpart
24.1199 protons 0.250000 seconds 0.119900 decpart
23.1922 protons 0.260000 seconds 0.192213 decpart
20.0999 protons 0.300000 seconds 0.099920 decpart
M =
πρ
0
R
2
3
R
s
R
s
=
3M
p
π(0.78Ge + 0.21Si )
1
α
2
m
p
h4π r
2
p
Gc
∫
3
2
1
t
2
dt = 0.78
∫
3
2
cos
−1
(x /2)d x = 0.21
air = 0.78N
2
+ 0.21O
2
air
H
2
O
≈ Φ
M
p
of 24 42
17.2285 protons 0.350000 seconds 0.228504 decpart
15.0749 protons 0.400000 seconds 0.074944 decpart
14.0232 protons 0.430000 seconds 0.023204 decpart
13.1086 protons 0.460000 seconds 0.108647 decpart
12.0600 protons 0.500000 seconds 0.059957 decpart
We see Germanium(32 protons) occurs between 0.15 and 0.20 seconds and silicon (12
protons) occurs at 0.5 seconds. Let’s hone the first.!
By what value would you like to increment?: 0.00355
How many values would you like to calculate for t in equation 1 (no
more than 100?): 100
566.1949 protons 0.010650 seconds 0.194885 decpart
283.0974 protons 0.021300 seconds 0.097443 decpart
113.2390 protons 0.053250 seconds 0.238968 decpart
106.1615 protons 0.056800 seconds 0.161537 decpart
77.2084 protons 0.078100 seconds 0.208389 decpart
53.0808 protons 0.113600 seconds 0.080769 decpart
47.1829 protons 0.127800 seconds 0.182907 decpart
36.1401 protons 0.166850 seconds 0.140114 decpart
32.0488 protons 0.188150 seconds 0.048786 decpart
And we see it occurs at 0.188150 seconds. At 47/250.!
We see if we take equation 3:!
!
Then where !
!
In Planck-seconds where !
We have:!
!
!
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon ds
1
α
2
m
p
h4π r
2
p
Gc
= (6protons)(1secon d )
5.391247(60)E − 44 second s = 1Planck Secon d
1
α
2
m
p
h4π r
2
p
Gc
= 6proton ⋅ secon ds
1
α
2
m
p
h4π r
2
p
Gc
= 1E 43proton ⋅ planck ⋅ secon d s
of 25 42
Si=9.274291168E42 planck seconds!
Ge=3.487133479E42 planck seconds!
12Si=1E44 proton planck seconds!
32Ge=1E44 proton planck seconds!
Where 12 and 32 are the protons in Si and Ge respectively. Concerning equations 7, 8, 9,10,
11, and indeed equation 1, we should be interesting in the ratio 78% and 21% and the way
they work with one another.!
0.78=39/50!
0.21=21/100!
Which are basically and .!
Planets Like An Atom
In so much as we have a definition for the radius of the solar system!
!
!
!
Through life at its fundamental structure the hydrocarbons!
!
!
Which occurs in air and water that makes life possible!
!
!
π /4
(1 − π /4)
R
s
=
3M
p
π(0.78Ge + 0.21Si )
1
α
2
m
p
h4π r
2
p
Gc
∫
3
2
1
t
2
dt = 0.78
∫
3
2
cos
−1
(x /2)d x = 0.21
1
t
1
1
α
2
m
p
h4π r
2
p
Gc
= 6protons
1
t
6
1
α
2
m
p
h4π r
2
p
Gc
= 1proton
air = 0.78N
2
+ 0.21O
2
air
H
2
O
≈ Φ
of 26 42
We want to consider its ground state, and because Mercury the first planet is so small and
carries little energy we proceed to Venus and describe it in terms of silicon and germanium like
we did with our definition for the radius of the solar system. I find by molar mass we have its
average orbital distance from the sun (0.72 AU) precisely in the following expression:!
Equation 12. !
Where Si=28.09 grams/mole and Ge=72.64 grams/mole. If we rewrite this:!
Equation 13. !
And differentiate!
!
!
And evaluate these for silicon and germanium we have!
!
!
Which is:!
!
Thus doing as we did before, we move down group 14 of the periodic table from carbon, to
silicon, to germanium and have:!
C+Si+Ge=12.01+28.09+72.64=112.74g/mol!
We have!
!
0.72AU =
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
r(x, y) =
1
y
2
2x y +
x
3
y
1 +
x
2
y
2
∂r
∂x
=
x
4
+ x
2
y
2
+ 2y
4
y(x
2
+ y
2
)
2
∂r
∂y
=
x(x
4
+ x
2
y
2
+ 2y
4
)
y
2
(x
2
+ y
2
)
2
∂r
∂x
= 0.0022626491
∂r
∂y
= 0.008749699
(0.00226)
2
+ (0.00875)
2
= 0.009037151
AU ⋅ m ol
gram
0.009037
AU ⋅ m ol
g
⋅ 112.74
g
mol
= 1.0AU
of 27 42
Which is Earth Orbit. Plots of our equation for Venus are:!
!
It may be we want to consider Jupiter the ground state as it carries the majority of the mass of
the solar system. We may have that Venus is the ground state for the terrestrials planets and
Jupiter the ground state for the outer gas giants, as separated by the asteroid belt, the location
where a planet cannot form. This takes us to Mars, which is beyond which it lay."
of 28 42
Mars
The asteroid belt is about 1AU thick at 2.2 to 3.2 AU from the sun, the earth-sun separation
being 1 AU. Carbon-12 the basis of life as we know it is 12.01 g/mol. It is made in stars from
Beryllium-8 is 8.0053051 g/mol. Thus we have!
Equation 14. !
Equation 15. !
Delta x equal to 1 AU is both Earth and The Asteroids. Mars is at 1.52AU. Delta X cancels with
the Earth leaving Mars equal to carbon to beryllium, which is life, Does this say we need to be
able to colonize Mars to succeed as a species?!
!
Or, does it mean we need to put bases on the moon to mine Helium-3 as a clean, renewable
energy source.!
The Moon
Essentially as the moon orbits the earth it makes 12 revolutions for each revolution of the Earth
around the sun which is 365.25 days. That is to say!
!
!
These are frequencies of!
!
!
In radians per day these are:!
!
!
Thus the equations of their phases are:!
!
!
Where t is in days and 2.55605E-3=!
(radius lunar orbit)(radius earth orbit)=384,400km/(149,597,876km).!
12
C
8
Be
=
Mars
Ear th
Δx
Δx = 1AU
12
C =
4
He +
8
Be
T
e
= 365.25d ays
T
m
= 29.53059d ays
f
e
= 0.002737851s
−1
f
m
= 0.033863191days
−1
ω
e
= 0.0172
ω
m
= 0.21
y
e
= cos(ω
e
t)
y
m
= (2.57E − 3)cos(ω
m
t)
of 29 42
We can say the frequency of the moon is 0.21/0.0172=12.21 times greater than that of the
earth. Thus we have the following plots of lunar phases to earth phases:!
There are 12 moons in a year and 24 hours in a day. Divide twelve by 2 and we have 6, divide
24 by 2 and we have 12. We have:!
!
In that days=1, moons-12, hours=24.!
At this point we bring-up that remarkable fact the the moon perfectly eclipses the sun. This is
because:!
!
!
Which are approximately equal. As well we can look at it as:!
!
!
Which are about the same as well. The interesting thing is that since our ratios are around
0.0025 and 0.0045, then…!
moon s
2
hours ⋅ days
= 6
(lunar − orbit)
(ear th − orbit)
=
384,400k m
149,592,870k m
= 0.00257
(lunar − radius)
solar − ra dius
=
1,738.1
696,00
= 0.0025
(lunar − radius)
(lunar − orbit)
=
(1,738.1)
(384,400)
= 0.00452
solar − ra dius
ear th − orbit
=
696,000
149,597,870
= 0.00465
of 30 42
!
I say this is interesting because this the ratio of the precious metal gold (Au) to that of silver
(Ag) by molar mass:!
!
We have:!
!
!
Taking the average between these:!
Equation 16. !
Where, is the lunar orbit, is the solar radius, is the earth orbit, and is the radius of
the moon. What this means is that the moon perfectly eclipses the sun because the solar
radius is 1.8 times greater than the lunar orbital radius. And interestingly gold is yellow, silver is
silver and the sun is gold, and the moon is silver, the moon perfectly eclipses the sun allowing
us to study its outer atmosphere, and the exact number of moons in a year is:!
!
x=12.3685304!
Is the consecutive integers 0, 1, 2, 3, 4. 5, 6,…8 only missing the 7. Where the equation!
square moons per 24-hour day,…!
Is concerned, I believe it is deeply interacting throughout nature with the equation for proton-
seconds:!
!
Because !
0.0045
0.0025
=
9
5
= 1.8
Au
Ag
=
196.97
107.87
= 1.8
(lunar − radius)(earth − orbit)
(lunar − orbit)
2
= 1.759577590
(solar − ra dius)
2
(ear th − orbit)(lun ar − radius
= 1.863
1
2
⋅
(
r
2
m
⋅ R
2
⊙
+ r
2
e
R
2
m
r
2
m
⋅ r
e
⋅ R
m
)
=
Au
Ag
r
m
R
⊙
r
e
R
m
365.25
x
= 29.53059d ays
(12m oon s)
2
24hours ⋅ 1day
= 6
1
t
1
α
2
m
p
h4π r
2
p
Gc
= 6protons
t
1
= 1second
of 31 42
And there are 60 minutes in an hour, and 60 seconds in a minute. We can equate them
because while:!
!
Has units of mass, dividing it by gives a number of protons, but but as well you can think of
the cancelling with the mass leaving pure number, and in!
!
You can think of 12 moons being the number of new moons that appear in the sky per time of
the journey of the sun North and back South again marking the seasons. We can divide each
day into 24 units and so it may be that there is some dynamic behind:!
!
Meaning:!
!
Theoretical Value For Proton Radius
In that we have!
!
And the periodic table of the elements is cyclical with 18 groups and!
!
Then perhaps we are supposed to write!
!
In fact, what if the 3 is supposed to be pi, then!
1
t
1
1
α
2
⋅
h4π r
2
p
Gc
m
p
m
p
(12m oon s)
2
24hours ⋅ 1day
= 6
1
t
1
α
2
m
p
h4π r
2
p
Gc
=
(12m oon s)
2
24hours ⋅ 1day
6proton s =
(12m oon s)
2
24hours ⋅ 1day
= C
1
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
6 =
1
α
2
m
p
h4π r
2
p
Gc
3
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
of 32 42
!
Then we would say that!
k=18/pi=5.7229577951!
The parameter in our constant with the most uncertainty is the radius of a proton . If the 3 is
supposed to be pi, then the radius of a proton becomes:!
!
Which gives!
!
About 95% raw most recent value measured. But, if !
!
Is supposed to be 6 and it is supposed to be multiplied by three to give 18 even which we
need for chemistry so we have 18 protons in the last group of the periodic table which is
important because we need argon with 18 protons for predicting valence numbers of elements
in terms of their need to attain noble gas electron configuration. Then we get!
!
This is in very close agreement with the most recent value measured which is!
!
Which is good for chemistry.!
π
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
r
p
r
p
= k α
2
m
p
Gc
4πh
r
p
= 8.790587E − 16m
1
α
2
m
p
h4π r
2
p
Gc
r
p
= 8.288587E − 16m = 0.829f m
r
p
= 0.833 + / − 0.014
of 33 42
Part 3: The Wave-Packet!
of 34 42
Thus we are interested in:!
!
!
Because we see these elements are a gaussian distribution (See graph on next page) and a
wave graph (page 39)…!
!
π
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u ) ⋅ d
a =
5
∏
i=1
x
i
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ Ag ⋅ Au
of 35 42
"
of 36 42
We consider a Gaussian wave-packet at t=0:!
!
We say that d which in quantum mechanics would be the delocalization length when squared is
. A is the amplitude and we might say it is . We write the wave packet as a
Fourier transform which is:!
!
We use the identity that gives the integral of a quadratic:!
!
Solve the equation!
!
With the initial condition!
!
A plane wave is the solution:!
!
Where, !
The wave-packet evolves with time as!
!
Calculate the Gaussian integral of dp!
and !
The solution is:!
ψ (x,0) = Ae
−
x
2
2d
2
Si − C
C
Ga − P
C
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
h
px
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
iℏ∂
t
ψ (x, t) =
p
2m
ψ (x, t)
ψ (x,0) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
px
e
i
ℏ
( px−ϵ( p)t)
ϵ( p) =
p
2
2m
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
( px−
p
2
2m
t)
α =
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
of 37 42
where !
We notice here one of the things you can do with equation 6 of part 2 is integrate from 0.5 sec
to 1 sec and you get one which multiplied by the constant which is six yields six. Now look up
0.5 seconds from the data output from the program and it is silicon, then go to one second and it
is carbon, thus the integral from silicon in time to carbon in time is carbon. Now consider life as
we know it is based on carbon because it has four valence electrons, but it is not based on
silicon, which has four valence electrons as well, because in the presence of oxygen it readily
forms SiO2 (sand or glass) leaving it unavailable to nitrogen, phosphorus, and hydrogen to
make make amino acids the building blocks of life. But silicon can be doped with phosphorus,
boron, gallium, and arsenic to make semiconductors -- transistor technology from which we can
build artificial life (artificial intelligence, AI). We can integrate over many time ranges to explore
millions of more facets to the equation.
We take our solution for the probability!
!
And we say it is:!
Equation 1. !
Let’s say x=1 proton which corresponds to 6 seconds. Then the probability we have a proton is
100% or, the absolute value of psi squared equals 1 means one second meaning we have
carbon or, life in other words. But one it does equal:!
!
We should say!
or !
ψ
2
= ex p
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
1
α
2
m
p
h4π r
2
p
Gc
∫
t
C
t
Si
1
t
2
dt = 6 = carbon(C )
t
Si
= 0.5secon ds
t
C
= 1secon d
ψ
2
= ex p
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
ψ
2
= ex p
[
−
Cx
2
(Si − C )
⋅
1
1 + t
2
/τ
2
]
exp
[
3
4
⋅ (1proton)
2
⋅
1
1 + (6second s)
2
]
= 1.02
d =
Si − C
C
=
4
3
d
2
=
16
9
of 38 42
And is almost exactly one if we average it with the following:!
!
The way I am using equation 1 is . We have:!
Equation 2. !
Thus for hydrogen:!
!
=74%.!
For Helium:!
=26%!
This is interesting because the Universe is about 74% Hydrogen and 24% Helium, the rest of
the elements making up the other 2%. Thus we can say or . We have
multiplied the first by 1 for Hydrogen element 1, and the second by 2 for helium element 2. In a
sense then, the probabilities represent the probability of finding hydrogen and helium in the
Universe. Hydrogen and and much of the helium were made theoretically in the Big Bang of the
big bang theory. The other elements were synthesized from these by the stars. Now we move
on to the other set of data in:!
!
Which is an interesting type of wave enveloped by a couple straight lines with different slopes
(Next page)…"
exp
[
3
4
⋅ (1proton)
2
⋅
1
(6secon d s)
2
]
= 0.979382
τ = d
2
ψ
2
= ex p −
C
2
x
2
(Si − C )
2
⋅
1
1 +
[
ℏC
2
m(Si − C)
2
]
2
t
2
ψ
2
= ex p −
9
16
x
2
⋅
1
1 +
ℏ
2
81
m
2
256
t
2
ψ
2
= (1)ex p −
9
16
(1proton)
2
⋅
1
1 +
(0.075)81
(1)256
(6secon d s)
2
ψ
2
= (2)ex p −
9
16
(2proton)
2
⋅
1
1 +
(0.075)81
(4)256
(3secon d s)
2
ℏ
2
= 0.075
ℏ = 0.27386
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ Ag ⋅ Au
of 39 42
"
of 40 42
Thus the graph on page 35 refers to!
!
And the graph on page 39 refers to!
!
And the two are equal, which are respectively,…!
!
!
If the first is the probability of finding hydrogen and helium in the Universe determined by a
wave function, and the second is a wave, then…"
π
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u ) ⋅ d
a
5
∏
i=1
x
i
of 41 42
We say the area of the integral as approximated by triangles is:!
!
=1.3+0.38+2.933+1.595+2.676+5.1665=12.75!
We subtract 3(1.1338) and get 12.75-3.4=9.35!
We see that we have come full circle because!
!
Which are the number of groups in the periodic table.!
Indeed we have come full circle because!
Equation 3. !
Equation 4. !
Where and .!
∫
3
1
f (n)dn = A1 + A2 + A3 + R1 + R2 + R3
(
Au − Ag
C
−
Si − C
C
)
Δn = 18
3
(
Au − Ag
C
−
Si − C
C
)
= 18
3
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
t
1
= 0.5
t
2
= 1
of 42 42
The Author
